In a combustion of a hydrocarbon compound, 2 reactions are happening per element:
C + O₂ → CO₂
2 H + 1/2 O₂ → H₂O
Thus, we can determine the amount of C and H from the masses of CO₂ and H₂O produced, respectively.
1.) Compute for the amount of C in the compound. The data you need to know are the following:
Molar mass of C = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Solution:
0.5008 g CO₂*(1 mol CO₂/ 44 g)*(1 mol C/1 mol CO₂) = 0.01138 mol C
0.01138 mol C*(12 g/mol) = 0.13658 g C
Compute for the amount of H in the compound. The data you need to know are the following:
Molar mass of H = 1 g/mol
Molar mass of H₂O = 18 g/mol
Solution:
0.1282 g H₂O*(1 mol H₂O/ 18 g)*(2 mol H/1 mol H₂O) = 0.014244 mol H
0.014244 mol H*(1 g/mol) = 0.014244 g H
The percent composition of pure hydrocarbon would be:
Percent composition = (Mass of C + Mass of H)/(Mass of sample) * 100
Percent composition = (0.13658 g + 0.014244 g)/(<span>0.1510 g) * 100
</span>Percent composition = 99.88%
2. The empirical formula is determined by finding the ratio of the elements. From #1, the amounts of moles is:
Amount of C = 0.01138 mol
Amount of H = 0.014244 mol
Divide the least number between the two to each of their individual amounts:
C = 0.01138/0.01138 = 1
H = 0.014244/0.01138 = 1.25
The ratio should be a whole number. So, you multiple 4 to each of the ratios:
C = 1*4 = 4
H = 1.25*4 = 5
Thus, the empirical formula of the hydrocarbon is C₄H₅.
3. The molar mass of the empirical formula is
Molar mass = 4(12 g/mol) + 5(1 g/mol) = 53 g/mol
Divide this from the given molecular weight of 106 g/mol
106 g/mol / 53 g/mol = 2
Thus, you need to multiply 2 to the subscripts of the empirical formula.
Molecular Formula = C₈H₁₀
1.33 is the answer thank me later
Answer:
If temperature increases, as it does in most reactions, a chemical change is likely to be occurring. This is different from the physical temperature change. During a physical temperature change, one substance, such as water is being heated.
Explanation:
<span>An exothermic reaction is one in which heat is released from the reagents into the ambient environment. Perhaps somewhat counterintuitively, condensation is in fact an example of such a reaction. During the process of the gas-to-liquid phase change, water goes from a higher-energy to lower-energy state of matter, and, as such, releases heat into the environment.</span>
Answer:
NO2- is the reducing agent.
Cr2O7_2- is the oxidizing agent.
H+ is neither
Explanation:
Reduction is the gain in electron. A chemical specie that undergoes reduction is called the oxidizing agent.
Oxidation is simply the loss in electrons. A chemical specie that undergoes oxidation is called the reducing agent.
Let us look at the species.
The first specie is the NO2-. In this specie, the oxidation number of nitrogen changed from +3 to +5 in NO3-. Thus we can see that there is more loss of electron to have caused an increase in the oxidation number positively. This shows an oxidation. Hence, NO2- is the reducing agent.
Let us look at the chromium. We can see that the oxidation number of chromium changed from +7 to +3.
Now we can see that it is a decrease and hence, it is a gain of electron and thus it is reduction. This means the first chromium specie is the oxidizing agent.
The hydrogen ion is simply placed there to balance the ions and hence it is neither the oxidizing nor the reducing agent.
